Question

我想在python中编写一个League Fixture生成器，但我不能。以下是详细信息：

有一个像teams = ["Team1", "Team2", "Team3", "Team4"]这样的团队的动态列表。如何从团队列表中生成fixture_weekx列表？例如：

fixture_week1 = ["Team1", "Team2", "Team3", "Team4"]
fixture_week2 = ["Team1", "Team3", "Team2", "Team4"]
fixture_week2 = ["Team1", "Team4", "Team2", "Team3"]

#Return matches:
fixture_week1 = ["Team2", "Team1", "Team4", "Team3"]
fixture_week2 = ["Team3", "Team1", "Team4", "Team2"]
fixture_week2 = ["Team4", "Team1", "Team3", "Team2"]

有什么想法吗？

Answer 1

夹具调度是一个众所周知的问题。这是在http://en.wikipedia.org/wiki/Round-robin_tournament

中给出的算法的python实现

# generation code - for cut and paste

import operator
def fixtures(teams):
    if len(teams) % 2:
        teams.append('Day off')  # if team number is odd - use 'day off' as fake team     

    rotation = list(teams)       # copy the list

    fixtures = []
    for i in range(0, len(teams)-1):
        fixtures.append(rotation)
        rotation = [rotation[0]] + [rotation[-1]] + rotation[1:-1]

    return fixtures

# demo code
teams = ["Team1", "Team2", "Team3", "Team4", "Team5"]

# for one match each - use this block only
matches = fixtures(teams)
for f in matches:    
    print zip(*[iter(f)]*2)

# if you want return matches 
reverse_teams =  [list(x) for x in zip(teams[1::2], teams[::2])]
reverse_teams = reduce(operator.add,  reverse_teams)    # swap team1 with team2, and so on ....

#then run the fixtures again
matches = fixtures(reverse_teams)

print "return matches"
for f in matches:    
    print f

这会生成输出：

[('Team1', 'Day off'), ('Team2', 'Team5'), ('Team3', 'Team4')]
[('Team1', 'Team5'), ('Day off', 'Team4'), ('Team2', 'Team3')]
[('Team1', 'Team4'), ('Team5', 'Team3'), ('Day off', 'Team2')]
[('Team1', 'Team3'), ('Team4', 'Team2'), ('Team5', 'Day off')]
[('Team1', 'Team2'), ('Team3', 'Day off'), ('Team4', 'Team5')]

Answer 2

我想评论来自@MariaZverina的代码不太有用。我按原样尝试了，但是没有得到正确的配对。我在下面做的修改与她的代码一起使用。不同之处在于我通过拉紧夹具f的前半部分来对每个夹具进行彩虹式配对与下半场相反。

# demo code
teams = ["Team1", "Team2", "Team3", "Team4", "Team5"]

# for one match each - use this block only
matches = fixtures(teams)    
for f in matches:    
    # This is where the difference is.
    # I implemented "rainbow" style pairing from each fixture f 
    # In other words: 
    # [(f[0],[f[n-1]), (f[1],f[n-2]), ..., (f[n/2-1],f[n/2])], 
    # where n is the length of f
    n = len(f)
    print zip(f[0:n/2],reversed(f[n/2:n]))

Answer 3

@MariaZverina的代码没有用，我也使用Round-robin_tournament实现了这段代码：

teams = ["Team1", "Team2", "Team3", "Team4", "Team5", "Team6"]
if len(teams) % 2:
    teams.append('Day off')
n = len(teams)
matchs = []
fixtures = []
return_matchs = []
for fixture in range(1, n):
    for i in range(n/2):
        matchs.append((teams[i], teams[n - 1 - i]))
        return_matchs.append((teams[n - 1 - i], teams[i]))
    teams.insert(1, teams.pop())
    fixtures.insert(len(fixtures)/2, matchs)
    fixtures.append(return_matchs)
    matchs = []
    return_matchs = []

for fixture in fixtures:
    print fixture

输出：

[('Team1', 'Team6'), ('Team2', 'Team5'), ('Team3', 'Team4')]
[('Team1', 'Team5'), ('Team6', 'Team4'), ('Team2', 'Team3')]
[('Team1', 'Team4'), ('Team5', 'Team3'), ('Team6', 'Team2')]
[('Team1', 'Team3'), ('Team4', 'Team2'), ('Team5', 'Team6')]
[('Team1', 'Team2'), ('Team3', 'Team6'), ('Team4', 'Team5')]
[('Team6', 'Team1'), ('Team5', 'Team2'), ('Team4', 'Team3')]
[('Team5', 'Team1'), ('Team4', 'Team6'), ('Team3', 'Team2')]
[('Team4', 'Team1'), ('Team3', 'Team5'), ('Team2', 'Team6')]
[('Team3', 'Team1'), ('Team2', 'Team4'), ('Team6', 'Team5')]
[('Team2', 'Team1'), ('Team6', 'Team3'), ('Team5', 'Team4')]

在python中的联盟夹具生成器

3 个答案: